MCQ
If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary, then the minimum value of $\mid \mathrm{z}-(3+3 \mathrm{i}) \mid$ is :
- A$2 \sqrt{2}-1$
- B$3 \sqrt{2}$
- C$6 \sqrt{2}$
- ✓$2 \sqrt{2}$
✓
Answer
Correct option: D.
$2 \sqrt{2}$
d
$\frac{z-i}{z-1}$ is purely Imaginary number
$\frac{z-i}{z-1}$ is purely Imaginary number
Let $z=x+$ iy
$\therefore \frac{x+i(y-1)}{(x-1)+i(y)} \times \frac{(x-1)-i y}{(x-1)-i y}$
$\Rightarrow \frac{x(x-1)+y(y-1)+i(-y-x+1)}{(x-1)^{2}+y^{2}}$ is purely
Imaginary number
$\Rightarrow \mathrm{x}(\mathrm{x}-1)+\mathrm{y}(\mathrm{y}-1)=0$
$\Rightarrow\left(\mathrm{x}-\frac{1}{2}\right)^{2}+\left(\mathrm{y}-\frac{1}{2}\right)^{2}=\frac{1}{2}$
$\therefore|z-(3+3 i)|_{\min } =\mid \mathrm{PCl}-\frac{1}{\sqrt{2}}$
$\quad\quad\quad\quad\quad\quad\quad\quad=\frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2 \sqrt{2}$
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